# On a divergent integral

## Main Question or Discussion Point

greetings . we have the integral :
$$\lim_{T\to \infty }\int_{2-iT}^{2+iT}\frac{(s-1)^{n}}{s}ds$$

which diverges for every value of n except $n=0$
if we perform the change of variables :

$$s\rightarrow \frac{1}{s}$$

then :
$$\lim_{T\to \infty }\int_{2-iT}^{2+iT}\frac{(s-1)^{n}}{s}ds=\int_{-i}^{i}\frac{(1-s)^{n}}{s^{n+1}}ds$$

which converges . am i missing something here , or is this correct !?

## Answers and Replies

greetings . we have the integral :
$$\lim_{T\to \infty }\int_{2-iT}^{2+iT}\frac{(s-1)^{n}}{s}ds$$

which diverges for every value of n except $n=0$
if we perform the change of variables :
Can you solve it for n=2 using antiderivatives? That is, what is:

$$\lim_{T\to\infty}\left(-2s+\frac{s^2}{2}+\log(s)\right)_{2-iT}^{2+iT}$$

mathman
Science Advisor
greetings . we have the integral :
$$\lim_{T\to \infty }\int_{2-iT}^{2+iT}\frac{(s-1)^{n}}{s}ds$$

which diverges for every value of n except $n=0$
if we perform the change of variables :

$$s\rightarrow \frac{1}{s}$$

then :
$$\lim_{T\to \infty }\int_{2-iT}^{2+iT}\frac{(s-1)^{n}}{s}ds=\int_{-i}^{i}\frac{(1-s)^{n}}{s^{n+1}}ds$$

which converges . am i missing something here , or is this correct !?
You need to split the original integration range into two parts at s = 0. Now when you change s to 1/s, you will be able to get the correct integration limits. Note also that the original T becomes 1/T.